Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, p\neq 0$. $\dfrac{{(k)^{-2}}}{{(k^{-3}p)^{-4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${k}$ to the exponent ${-2}$ . Now ${1 \times -2 = -2}$ , so ${(k)^{-2} = k^{-2}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-3}p)^{-4} = (k^{-3})^{-4}(p)^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k)^{-2}}}{{(k^{-3}p)^{-4}}} = \dfrac{{k^{-2}}}{{k^{12}p^{-4}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-2}}}{{k^{12}p^{-4}}} = \dfrac{{k^{-2}}}{{k^{12}}} \cdot \dfrac{{1}}{{p^{-4}}} = k^{{-2} - {12}} \cdot p^{- {(-4)}} = k^{-14}p^{4}$.